Chapter 7.3, Problem 47E

Solution

To calculate: The geometric sequence that contains the two specified terms

  $a_4=6\text{and}{a}_7=\frac{243}{8}$

The required rule is $a_n=\frac{32}{27}\cdot {\left[\left(\frac{81}{27}\right)\right]}^{\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Given information:

The given terms are $a_4=6\text{and}{a}_7=\frac{243}{8}$ .

Formula used:

The nth term of a geometric sequence with first term $a_1$ and common ratio $r$ is given by: $a_n=a_1\cdot r^{n-1}$

Calculation:

Consider the given sequence.

It is given that $a_4=6\text{and}{a}_7=\frac{243}{8}$ .

Geometric series have a common ratio $r$ , where $a_{n+1}=a_{nr}$

  $\begin{array}{l}{a}_4=6\\ a_5=6r\\ a_6=6r^2\\ a_7=6r^3\end{array}$

Determine the value of $r$ using this term $a_7=6r^3$

  $\begin{array}{c}6r^3=\frac{243}{8}\\ r^3=\frac{243}{6\cdot 8}\\ r^3=\frac{81}{16}\\ r=\sqrt[3]{\frac{81}{16}}\\ r\approx 1.717\end{array}$

Determine the $n=1$ term.

  $\begin{array}{l}{a}_3=\frac{6}{r}\\ a_2=\frac{6}{r^2}\\ a_1=\frac{6}{r^3}\end{array}$

Put the value of r into $\frac{6}{r^3}$

  $\begin{array}{l}{a}_1=\frac{6}{r^3}\\ a_1=\frac{6}{\frac{81}{16}}\\ a_1=\frac{96}{81}\\ a_1=\frac{32}{27}\end{array}$

Put the value of $a_1=\frac{32}{27}\text{and}r=\frac{81}{16}$ into $a_n=a_1{r}^{n-1}$

  $a_n=\frac{32}{27}\cdot {\left[\left(\frac{81}{27}\right)\right]}^{\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

There is solution.

  $a_n=\frac{32}{27}\cdot {\left[\left(\frac{81}{27}\right)\right]}^{\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$